Exercise 6.1.18

If ${⟨\text{⋅,⋅}⟩}^{\prime }$ is an inner product on $V$, then we have that $T(x)=0$ implies

and $x=0$. So $T$ is injective. Conversely, if $T$ is injective, we may check those condition for inner product one by one.

• $${⟨x+z,y⟩}^{\prime }=⟨T(x+z),T(y)⟩=⟨T(x)+T(z),T(y)⟩$$

  $$=⟨T(x),T(y)⟩+⟨T(z),T(y)⟩={⟨x,y⟩}^{\prime }+{⟨z,y⟩}^{\prime }.$$

• $${⟨\mathit{cx},y⟩}^{\prime }=⟨T(\mathit{cx}),T(y)⟩=⟨\mathit{cT}(x),T(y)⟩$$

  $$=c⟨T(x),T(y)⟩={⟨x,y⟩}^{\prime }.$$

• $$\overline{{⟨x,y⟩}^{\prime }}=\overline{⟨T(x),T(y)⟩}$$

  $$=⟨T(y),T(x)⟩={⟨y,x⟩}^{\prime }.$$

• $${⟨x,x⟩}^{\prime }=⟨T(x),T(x)⟩>0$$

  if $T(x)\ne 0$. And the condition $T(x)\ne 0$ is true when $x\ne 0$ since $T$ is injective.